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$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to
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$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$
Since, $\sin x \in[-1,1]$
Hence, when $x \rightarrow \infty$, then the function is of $\frac{\text { finite }}{\infty}$ form
$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right) \\ &= \frac{2+\text { value in between }(-1) \text { and }(1)}{\infty+3}=0\end{aligned}$
Since, $\sin x \in[-1,1]$
Hence, when $x \rightarrow \infty$, then the function is of $\frac{\text { finite }}{\infty}$ form
$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right) \\ &= \frac{2+\text { value in between }(-1) \text { and }(1)}{\infty+3}=0\end{aligned}$
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